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MDCAT Physics QUESTION #4395
Question 1
If the horizontal range of a projectile becomes half of its maximum possible horizontal range, the probable angle of projection is:
  • $15°$✔️
  • $30°$
  • $45°$
  • $60°$
Correct Answer Explanation
Maximum range occurs at $45°$: $R_{max} = \dfrac{v^2}{g}$. Range at angle $\theta$: $R = \dfrac{v^2\sin2\theta}{g}$. For $R = \dfrac{R_{max}}{2}$: $\sin2\theta = \dfrac{1}{2}$, so $2\theta = 30°$ or $150°$, giving $\theta = 15°$ or $75°$. The smallest angle is $15°$.